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Brillhart's Cubic and e^(Sqrt[163] π)

Posted 3 years ago
POSTED BY: Ed Pegg
2 Replies

Another trick

Brillhart = {{0, 2, 2}, {1, 0, 2}, {2, 1, 0}};
bc = Eigenvalues[Brillhart][[1]]; 
N[Log[(2 + bc)^24 - 24]/Sqrt[163], 34]

3.141592653589793238462643383279503

N[Pi, 34]

3.141592653589793238462643383279503

POSTED BY: Ed Pegg

Ed, you might what to add this to the entry in the OEIS in the PROG section at least.

https://oeis.org/A002937

The 163 number is treated in

https://www.youtube.com/watch?v=DRxAVA6gYMM

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